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Zorluk: ZorOrder of Operations and Number Properties

For all real numbers aa, bb, and cc, let the operation ⊕\oplus be defined by a⊕b=ab+1a \oplus b = ab + 1. Which of the following statements must be true?

I. a⊕b=b⊕aa \oplus b = b \oplus a
II. a⊕(b⊕c)=(a⊕b)⊕ca \oplus (b \oplus c) = (a \oplus b) \oplus c
III. a⊕(b+c)=(a⊕b)+(a⊕c)a \oplus (b + c) = (a \oplus b) + (a \oplus c)

  1. I onlyCevap
  2. B
    II only
  3. C
    I and II only
  4. D
    I and III only
  5. E
    I, II, and III

Cevap

The correct option is the one stating that only Statement I must be true.
The correct option is the one stating that only Statement I must be true. This is because multiplication of real numbers is commutative, making the custom operation commutative. Statement II is false because the operation is not associative, and Statement III is false because the operation does not distribute over addition.

Adım Adım Çözüm

1
Evaluate Statement I (a⊕b=b⊕aa \oplus b = b \oplus a) by substituting the definition of the operation.
a⊕b=ab+1a \oplus b = ab + 1 and b⊕a=ba+1b \oplus a = ba + 1. Since multiplication of real numbers is commutative, ab=baab = ba, which means ab+1=ba+1ab + 1 = ba + 1. Thus, Statement I is true for all real numbers.
To test the commutative property of the defined operation.
2
Evaluate Statement II (a⊕(b⊕c)=(a⊕b)⊕ca \oplus (b \oplus c) = (a \oplus b) \oplus c) by applying the operation's definition to both sides.
Left side: a⊕(b⊕c)=a⊕(bc+1)=a(bc+1)+1=abc+a+1a \oplus (b \oplus c) = a \oplus (bc + 1) = a(bc + 1) + 1 = abc + a + 1. Right side: (a⊕b)⊕c=(ab+1)⊕c=(ab+1)c+1=abc+c+1(a \oplus b) \oplus c = (ab + 1) \oplus c = (ab + 1)c + 1 = abc + c + 1. Since abc+a+1abc + a + 1 is not equal to abc+c+1abc + c + 1 for all real numbers (for example, if a=1a = 1, b=1b = 1, and c=2c = 2, then 4≠54 \neq 5), Statement II is not always true.
To test the associative property of the defined operation.
3
Evaluate Statement III (a⊕(b+c)=(a⊕b)+(a⊕c)a \oplus (b + c) = (a \oplus b) + (a \oplus c)) using the operation's definition.
Left side: a⊕(b+c)=a(b+c)+1=ab+ac+1a \oplus (b + c) = a(b + c) + 1 = ab + ac + 1. Right side: (a⊕b)+(a⊕c)=(ab+1)+(ac+1)=ab+ac+2(a \oplus b) + (a \oplus c) = (ab + 1) + (ac + 1) = ab + ac + 2. Since ab+ac+1≠ab+ac+2ab + ac + 1 \neq ab + ac + 2 for all real numbers, Statement III is false.
To test whether the operation distributes over addition.

Anahtar Kavram

Identifying properties of operations (commutativity, associativity, and distributivity) on real numbers.

Alternatif Yöntem

To quickly disprove Statements II and III, you can substitute simple counterexamples. For Statement II, let a=1,b=1,c=2a=1, b=1, c=2: 1⊕(1⊕2)=1⊕3=41 \oplus (1 \oplus 2) = 1 \oplus 3 = 4, while (1⊕1)⊕2=2⊕2=5(1 \oplus 1) \oplus 2 = 2 \oplus 2 = 5. Since 4≠54 \neq 5, Statement II is false. For Statement III, let a=1,b=1,c=1a=1, b=1, c=1: 1⊕(1+1)=1⊕2=31 \oplus (1+1) = 1 \oplus 2 = 3, while (1⊕1)+(1⊕1)=2+2=4(1 \oplus 1) + (1 \oplus 1) = 2 + 2 = 4. Since 3≠43 \neq 4, Statement III is false.
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Konular

Order of Operations and Number Properties

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